Question 6

We will use the concept of areas of triangles to prove this.

Step 1 — Draw altitudes and find areas

Let's draw a line segment $EF$ from E perpendicular to AB. So, $EF$ is the height for $\triangle ADE$ and $\triangle BDE$ when their bases are on AB.

The area of $\triangle ADE$ is:

$\text{Area}(\triangle ADE) = \frac{1}{2} \times \text{AD} \times \text{EF}$

The area of $\triangle BDE$ is:

$\text{Area}(\triangle BDE) = \frac{1}{2} \times \text{DB} \times \text{EF}$

Now, let's draw a line segment $DG$ from D perpendicular to AC. So, $DG$ is the height for $\triangle ADE$ and $\triangle CDE$ when their bases are on AC.

The area of $\triangle ADE$ is also:

$\text{Area}(\triangle ADE) = \frac{1}{2} \times \text{AE} \times \text{DG}$

The area of $\triangle CDE$ is:

$\text{Area}(\triangle CDE) = \frac{1}{2} \times \text{EC} \times \text{DG}$

Step 2 — Form ratios and conclude

Let's find the ratio of the areas of $\triangle ADE$ and $\triangle BDE$.

$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times \text{AD} \times \text{EF}}{\frac{1}{2} \times \text{DB} \times \text{EF}}$

$= \frac{\text{AD}}{\text{DB}}$

Next, let's find the ratio of the areas of $\triangle ADE$ and $\triangle CDE$.

$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times \text{AE} \times \text{DG}}{\frac{1}{2} \times \text{EC} \times \text{DG}}$

$= \frac{\text{AE}}{\text{EC}}$

Now, consider $\triangle BDE$ and $\triangle CDE$. They share the same base, which is DE. They are between the same parallel lines, DE and BC. Triangles on the same base and between the same parallel lines have equal areas. So, we can say:

$\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE)$

Since Area($\triangle BDE$) equals Area($\triangle CDE$), we can equate the two ratios we found earlier.

$\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$

Answer

(i) We have proven that $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$.